- Complete a one-sample [latex]t[/latex]-test for means from hypotheses to conclusions.
In the previous page, we looked at the following scenario:
Researchers want to examine the effect of diet on cholesterol levels. They select a random sample of [latex]125[/latex] adult males who are vegetarians and test their cholesterol levels to determine if they are significantly different from [latex]201.5[/latex] milligrams of cholesterol per deciliter (mg/dl), which is the average cholesterol level of all adult males without heart disease. We have found that the conditions to conduct a one-sample hypothesis test have been satisfied.
Use the following statistical tool to find the test statistic, and let’s use it to make an inference about the population.
Step 1: Under Enter Data, select Summary Statistics.
Step 2: The sample results about the cholesterol level are shown in the following table. Enter the sample results accordingly.
| Size | Mean | Standard Deviation | |
| Sample | [latex]125[/latex] | [latex]183.4[/latex] | [latex]15[/latex] |
Step 3: Under Type of Inference, select Significance Test.
Step 4: Enter the null value (in this scenario is [latex]201.5[/latex]) and select the correct alternative hypothesis.
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Notice that in the statistical tool above, the test statistic we are using is the [latex]t[/latex]-statistic. Recall that in practice, we rarely know the population standard deviation. So, we use the sample standard deviation [latex]s[/latex] as an estimate for [latex]\sigma[/latex]. Because of this estimation, we now use the [latex]t[/latex]-distribution instead of the normal distribution.
[latex]t=\dfrac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}[/latex]
For each sample size [latex]n[/latex], there is a different [latex]t[/latex]-distribution, which depends on the sample’s degree of freedom, [latex]df=n-1[/latex].